Determinant of $(I+uv^*)$ Let $u,v \in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?
 A: Assuming $uv^*\ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $\det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
$$
\det(uv^*+I)=v^*u+1
$$
A: Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
$$
{\displaystyle \det \left({A} +{uv} ^{\textsf {T}}\right)
=\left(1+{v} ^{\textsf {T}}{A} ^{-1}{u} \right)\,\det \left({A} \right)\,.}
$$
In your case, let $A=I$.
The proof is essentially based on the following observation:
$$
{\displaystyle {\begin{pmatrix}\mathbf {I} &0\\\mathbf {v} ^{\textsf {T}}&1\end{pmatrix}}{\begin{pmatrix}\mathbf {I} +\mathbf {uv} ^{\textsf {T}}&\mathbf {u} \\0&1\end{pmatrix}}{\begin{pmatrix}\mathbf {I} &0\\-\mathbf {v} ^{\textsf {T}}&1\end{pmatrix}}={\begin{pmatrix}\mathbf {I} &\mathbf {u} \\0&1+\mathbf {v} ^{\textsf {T}}\mathbf {u} \end{pmatrix}}.}
$$

[Added later:]
Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{\textsf {T}}$ is a rank one matrix. Moreover, $\lambda = {v} ^{\textsf {T}}u$ is an eigenvalue of $B$ since:
$$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
$$
(v^Tu,0,\cdots,0).
$$
It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
$$
(1+v^Tu,1,\cdots,1).
$$
But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.
1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$
A: You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u\,.$$
